Question

Find 3 numbers in ap whose sum is 18 and product is 192

Answer 1

Heya !!



Let the required numbers ate ( a - d ) , a , ( a + d ).




Then,



a - d + a + a + d = 18


3a = 18.


a = 6.



And,


( a - d ) a ( a + d ) = 192





=> a ( a² - d² ) = 192



=> 6 ( 36 - d² ) = 192




=> 6d² = 216 - 192



=> 6d² = 24



=> d² = 4



=> d = √4 = 2


First term ( a ) = 6


And,


Common difference (d ) = 2.






Hence,


The required numbers are ( 4 , 6 , 8)

Answer 2

Hey !

let three Numbers are ( a - d ) , a and (a + d ).

a - d - a + a + d = 18

3a = 18

a = 18/3 = 6

a = 6

and,

( a - d ) a ( a + d ) = 192

a ( a - d ) ( a + d ) = 192

a ( a² - d²) = 192

6 ( 36 - d² ) = 192

-6d² = -24

d² = 4

d = √4 = 2.

Hence , required numbers are 4 , 6 and 8.